Ultimate Load

w_u = 1.2 DL + 1.6 LL = 1.2×(0.75) + 1.6×(1.0) = 2.5 kip/ft

Slenderness

Slenderness ratio: = $\raisebox{1ex}{$L$}\!\left/ \!\raisebox{-1ex}{$r_y$}\right.=\frac{180}{0.865}=208.1$

Compare this to the allowable tension slenderness ratio of 300 (note that no tension or compression is present in the member):

ratio = 208.1 / 300 = 0.694, OK

Section Classification

Compression elements

• Flange classification per Table B4.1a, Case 1:
  $\lambda =\frac{b_f}{t_f}=\frac{3.72}{0.65}=5.723<{\lambda }_r=0.56\sqrt{\frac{E}{F_y}}=0.56\sqrt{\frac{\mathrm{29,000}}{36}}=15.89$

  thus, flanges are non-slender

• Web classification per Table B4.1a, Case 5:
  $\lambda =\frac{\left(d-2t_f\right)}{t_w}=\frac{\left(15-2\times 065\right)}{0.716}=19.13<{\lambda }_{\mathrm{wr}}=1.49\sqrt{\frac{E}{F_y}}=1.49\sqrt{\frac{\mathrm{29,000}}{36}}=42.30$

  thus, web is non-slender

Flexure elements

• Flange classification per Table B4.1b, Case 10:
  $\lambda =\frac{b_f}{t_f}=5.723<{\lambda }_p=0.38\sqrt{\frac{E}{F_y}}=0.38\sqrt{\frac{\mathrm{29,000}}{36}}=10.79$

  thus, flanges are compact

• Web classification per Table B4.1b, Case 15:
  $\lambda =\frac{\left(d-2t_f\right)}{t_w}=19.13<{\lambda }_{\mathrm{wp}}=3.76\sqrt{\frac{E}{F_y}}=3.76\sqrt{\frac{\mathrm{29,000}}{36}}=106.7$

  thus, web is compact

Tension Capacity

• Tensile Yielding

  Nominal tensile yield strength:

  PnY = FyAg = 36×14.7 = 529.2 kip

  (Eqn. D2-1)

  ϕtYPnY = 0.9×529.2 = 476.3 kip

• Tensile Rupture

  Effective area is equal to the net area, A_e = A_n = A_g = 14.7 in² (no bolt holes)

  Nominal tensile rupture strength:

  PnR = FuAe = 58×14.7 = 852.6 kip

  (Eqn. D2-2)

  ϕtRPnR = 0.75×852.6 = 639.5 kip

Compression capacity

• Flexural buckling about the X axis:

  the effective slenderness ratio about the local x axis (clause E2):

  $\frac{L}{r_x}=\frac{180}{5.24}=34.35<4.71\sqrt{\frac{E}{F_y}}=4.71\sqrt{\frac{\mathrm{29,000}}{36}}=133.7$

  The elastic buckling stress:

  $F_{\mathrm{ex}}=\frac{{\pi }^{2}E}{{\left(\raisebox{1ex}{$\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{$r_x$}\right.\right)}^2}=\frac{{\pi }^2\mathrm{29,000}}{{\left(34.35\right)}^2}=242.6\text{ ksi}$

  (Eqn. E3-4)

  Check that $\frac{F_y}{F_{\mathrm{ex}}}=\frac{36}{242.6}=0.148<2.25$

  $F_{\mathrm{crx}}={0.658}^{\raisebox{1ex}{$F_y$}\!\left/ \!\raisebox{-1ex}{$F_{\mathrm{ex}}$}\right.}\times F_y={0.658}^{0.148}\times 36=33.83\text{ ksi}$

  (Eqn. E3-2)

  Nominal compression capacity about the X axis:

  Pnc = FcrxAg = 33.83×14.7 = 497.3 kip

  (Eqn. E3-1)

  ϕcPnc = 0.9×497.3 = 447.6 kip

• Flexural buckling about the Y axis:

  the effective slenderness ratio about the local y axis (clause E2):

  $\frac{L}{r_y}=\frac{180}{0.865}=208.1>4.71\sqrt{\frac{E}{F_y}}=4.71\sqrt{\frac{\mathrm{29,000}}{36}}=133.7$

  The elastic buckling stress:

  $F_{\mathrm{ey}}=\frac{{\pi }^{2}E}{{\left(\raisebox{1ex}{$\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{$r_y$}\right.\right)}^2}=\frac{{\pi }^2\mathrm{29,000}}{{\left(208.1\right)}^2}=6.610\text{ ksi}$

  (Eqn. E3-4)

  Check that $\frac{F_y}{F_{\mathrm{ey}}}=\frac{36}{6.610}=5.47>2.25$

  $F_{\mathrm{cry}}=0.877F_{\mathrm{ey}}=0.877\times 6.610=5.80\text{ ksi}$

  (Eqn. E3-3)

  Nominal compression capacity about the Y axis:

  Pnc = FcryAg = 5.80×14.7 = 85.21 kip

  (Eqn. E3-1)

  ϕcPnc = 0.9×85.21 = 76.70 kip

• Flexural torsional buckling:
  Channel section is singly symmetric:

  • x₀ = e₀ + x = 0.583 + 0.490 = 1.382 in.
  • y₀ = 0

  Euler buckling stress:

  $F_{\mathrm{ez}}=\left(\frac{{\pi }^{2}E{C}_w}{L^2}+GJ\right)\frac{1}{A_g{\stackrel{\_}{r}}_0^2}=\left(\frac{{\pi }^2\left(\mathrm{29,000}\right)\left(492\right)}{{\left(180\right)}^2}+\mathrm{11,200}\times 2.65\right)\frac{1}{\mathrm{14.7}{\left(5.49\right)}^2}=76.80\text{ ksi}$

  (Eqn. E4-7)

  Elastic flexural torsional buckling stress:

  $F_e=\left(\frac{F_{\mathrm{ex}}+F_{\mathrm{ez}}}{2H}\right)\left[1-\sqrt{1-\frac{4F_{\mathrm{ex}}{F}_{\mathrm{ez}}H}{{\left(F_{\mathrm{ex}}+F_{\mathrm{ez}}\right)}^2}}\right]=\left(\frac{242.6+76.80}{2\left(0.937\right)}\right)\left[1-\sqrt{1-\frac{4\left(242.6\right)\left(76.80\right)\left(0.937\right)}{{\left(242.6+76.80\right)}^2}}\right]=74.71\text{ ksi}$

  (Eqn. E4-3)

  Critical flexural torsional buckling stress:

  Check that $\frac{F_y}{F_e}=\frac{36}{74.71}=0.482<2.25$

  $F_{\mathrm{cr}}={0.658}^{\raisebox{1ex}{$F_y$}\!\left/ \!\raisebox{-1ex}{$F_e$}\right.}\times F_y={0.658}^{0.482}\times 36=29.42\text{ ksi}$

  (Eqn. E3-2)

  Nominal compression capacity about the X axis:

  Pnc = FcrxAg = 29.42×14.7 = 432.5 kip

  (Eqn. E3-1)

  ϕcPnc = 0.9×432.5 = 389.3 kip

Shear Capacity

• Shear along the X axis ("weak axis" shear)

  Take k_v = 1.2 (clause G6)

  $\frac{b_f}{t_f}=5.723<1.10\sqrt{\frac{k_{v}E}{F_y}}=1.10\sqrt{\frac{1.2\times \mathrm{29,000}}{36}}=34.2$

  Therefore, C_v2 = 1.0 per clause G2.2 and the nominal shear strength in the X direction:

  $V_n=0.6F_{y}2b_f{t}_f{C}_{\mathrm{v2}}=0.6\left(36\right)2\left(3.72\right)\left(0.65\right)\left(1.0\right)=104.5\text{ kip}$

  (Eqn. G6-1)

  ϕvVn = 0.9×104.5 = 94.01 kip

• Shear along the Y axis ("strong axis" shear)

  Take k_v = 5.34 (clause G2.1.b.2.i)

  $\frac{\left(d-2t_f\right)}{t_w}=19.13<1.10\sqrt{\frac{k_{v}E}{F_y}}=1.10\sqrt{\frac{5.34\times \mathrm{29,000}}{36}}=72.15$

  Therefore, C_v1 = 1.0 per eqn. G2-3 and the nominal shear strength in the Y direction:

  $V_n=0.6F_{y}d{t}_w{C}_{\mathrm{v1}}=0.6\left(36\right)\left(15.0\right)\left(0.716\right)\left(1.0\right)=232.0\text{ kip}$

  (Eqn. G2-1)

  ϕvVn = 0.9×232.0 = 208.8 kip

Bending Capacity

• Flexural yielding about X axis

  $M_{\mathrm{nx}}=F_y{Z}_x=36\times 68.5=\mathrm{2,466}\text{ in·kip}=205.5\text{ ft·kip}$

  (Eqn. F2-1)

  ϕbMn = 0.9×205.5 = 185.0 ft·kip

• Flexural yielding about Y axis

  $M_{\mathrm{nx}}={}_{\text{min.}}{}^{}|\begin{array}{ll}{F}_y{Z}_y=36\times 8.14=293.0\text{ in·kip}=24.42\text{ ft·kip}& \\ 1.6F_y{S}_y=1.6\times 36\times 3.77=217.2\text{ in·kip}=18.10\text{ ft·kip}& \text{ ← governs}\end{array}_{}^{}$

  (Eqn. F6-1)

  ϕbMn = 0.9×18.10 = 16.29 ft·kip

• Lateral torsional buckling about X axis:

  The laterally unbraced length of the compression flange is 5 feet: L_b = 60 in.

  $L_p=1.76r_y\sqrt{\frac{E}{F_y}}=1.76\left(0.865\right)\sqrt{\frac{\mathrm{29,000}}{36}}=43.21\text{ in.}$

  (Eqn. F2-5)

  $L_r=1.95r_{\mathrm{ts}}\frac{E}{0.7F_y}\sqrt{\frac{Jc}{S_x{h}_0}+\sqrt{{\left(\frac{Jc}{S_x{h}_0}\right)}^2+6.76{\left(\frac{0.7F_y}{E}\right)}^2}}$

  (Eqn. F2-6)

  意味

  c

  =

  $\frac{h_0}{2}\sqrt{\frac{I_y}{C_w}}=\frac{14.4}{2}\sqrt{\frac{11.0}{492}}=1.077$ for channels

  $L_r=1.95\left(1.17\right)\frac{\mathrm{29,000}}{0.7\left(36\right)}\sqrt{\frac{2.65\times 1.077}{53.8\times 14.4}+\sqrt{{\left(\frac{2.65\times 1.077}{53.8\times 14.4}\right)}^2+6.76{\left(\frac{0.7\times 36}{\mathrm{29,000}}\right)}^2}}=234.9\text{ in.}$

  For L_p < L_b < L_r:

  $M_n=C_b\left[M_p-\left(M_p-0.7F_y{S}_x\right)\right]\left(\frac{L_b-L_p}{L_r-L_p}\right)$

  (Eqn. F2-2)

  意味

  Mp

  =

  Mn = 2,466 in·kips

  $=1.0\left[\mathrm{2,466}-\left(\mathrm{2,466}-0.7\times 36\times 53.8\right)\right]\left(\frac{60-43.21}{234.9-43.21}\right)=\mathrm{2,369}\text{ in·kip}=197.4\text{ ft·kip}$

  ϕbMn = 0.9×197.4 = 177.7 ft·kip

Combined Interaction Check

Since P_r = 0, P_r / P_c = 0 < 0.2, equation H1-1b governs: